I have this question from textbook A Course in Mathematical Analysis by Prof D. J. H. Garling at page 23. This exercise assumes only Zermelo–Fraenkel axioms of set, except for axiom of choice (which is presented in the next section). After some hours thinking, I still have no idea how to approach this exercise. Please shed some light on this problem. Many thanks!
Let $Z^{+}$ be the minimal infinite successor set. Consider the two-point subset $\{0,1\}$ of $Z^{+}$. We denote the set of functions from $A$ to $\{0,1\}$ by $2^{A}$. Suppose that $B\in P(A)$ and $x\in A$. Let $I_{B}(x)=1$ if $x\in B$, and $I_{B}(x)=0$ if $x\notin B$. $I_{B}$ is the indicator function of $B$. Show that the mapping $B\rightarrow I_{B}:P(A)\rightarrow 2^{A}$ is a bijection.
We are dealing here with a function $f:\wp(A)\to 2^A$ that is prescribed by $B\mapsto 1_B$.
In words: the function sends every element $B$ of $\wp(A)$ to the indicator function $1_B$ which on its turn is a function that has $A$ as domain and has $\{0,1\}$ as codomain.
Every $g\in 2^A$ induces a set $B_g:=\{a\in A\mid g(a)=1\}$ and it is not hard to show that: $$f(B_g)=1_{B_g}=g$$Proved is now that $f$ is surjective.
If $B,C\subseteq A$ with $B\neq C$ then $B-C\neq\varnothing$ or $C-B\neq\varnothing$.
Suppose that $B-C\neq\varnothing$ and let it be that $a\in B-C$.
Then $1_B(a)=1\neq0=1_C(a)$ so that $1_B\neq 1_C$.
Likewise it can be proved that $1_B\neq 1_C$ if $C-B\neq\varnothing$.
Proved is now: $$B\neq C\implies f(B)=1_B\neq1_C=f(C)$$which means exactly that $f$ is injective.
So $f$ is bijective and we are ready.